If $A,B,C$ are the angles of a triangle, what is the probability that $Ax+By \ge C$?

Question

Let $A,B,C$ be the angles of a triangle whose vertices are uniformly random on a circle. and let $x,y > 0$. Experimental data suggests that the probability $Ax + By > C$ has a simple closed form

$$ P(Ax + By > C) = \frac{xy + x + y}{xy + x + y + 1} = 1 - \frac{1}{(x+1)(y+1)} $$

Can this be proved or disproved?

Related question: If $(a,b,c)$ are the sides of a triangle, what is the probability that $ax+by \ge c$?

Update 12-Jun-2025: This problem can be geenralized as follows

Let $x_n> 0$ be a sequence of uniformly distributed reals whose sum is constant and $a_n > 0$; then,

$$ P\left(\sum_{r=1}^{n}a_r x_r \le x_{n+1}\right) = \prod_{r=1}^{n}\frac{1}{a_r+1} $$

Answer 1

To find the probability that $\text{A}x+\text{B}y\ge\text{C}$ for angles $\text{A},\text{B}$, and $\text{C}$ of a triangle, where $\text{A}+\text{B}+\text{C}=\pi$ and $x$ and $y$ are real numbers, we need to determine the proportion of all possible triangles (defined by their angles) for which this inequality holds. Since the angles are continuous and must satisfy $\text{A}>0$, $\text{B}>0$, $\text{C}>0$, and $\text{A}+\text{B}+\text{C}=\pi$, we interpret the probability as the ratio of the "area" of the region where the inequality is true to the total "area" of possible angle pairs, assuming a uniform distribution over the space of valid triangles.

First, let's define the space of possible triangles. The angles $\text{A},\text{B}$, and $\text{C}$ satisfy: $\text{A}+\text{B}+\text{C}=\pi$, $\text{A}>0$, $\text{B}>0$ and $\text{C}>0$. Since $\text{C}=\pi-\text{A}-\text{B}$, we can express $\text{C}>0$ as: $\pi-\text{A}-\text{B}>0\space\Longrightarrow\space\text{A}+\text{B}<\pi$. Thus, the possible values of $\text{A}$ and $\text{B}$ lie in the region: $\text{A}>0$, $\text{B}>0$, and $\text{A}+\text{B}<\pi$. This forms a right-angled triangle in the $\text{A}-\text{B}$ plane with vertices at $(0,0)$, $(\pi,0)$, and $(0,\pi)$, but constrained by $\text{A}+\text{B}<\pi$. The total area of this region represents all possible angle pairs $(\text{A},\text{B})$.

Now, lets find the total area. Consider the triangle with vertices $(0,0)$, $(\pi,0)$, and $(0,\pi)$. This is a right triangle with legs of length $\pi$. Its area is:

$$\text{Area}=\frac{1}{2}\cdot\pi\cdot\pi=\frac{\pi^2}{2}\tag1$$

However, the condition $\text{A}+\text{B}<\pi$ excludes the boundary line $\text{A}+\text{B}=\pi$ (where $\text{C}=0$), but in a continuous distribution, this boundary has measure zero and does not affect the area for integration purposes. Thus, the total area of the region is $\displaystyle\frac{\displaystyle\pi^2}{\displaystyle2}$.

Now, we can express the inequality. Substitute $\text{C}=\pi-\text{A}-\text{B}$ into the inequality $\text{A}x+\text{B}y\ge\text{C}$: $\text{A}x+\text{B}y\ge\pi-\text{A}-\text{B}$. Rearrange all terms: $\text{A}x+\text{B}y+\text{A}+\text{B}\ge\pi$, $\text{A}\left(x+1\right)+\text{B}\left(y+1\right)\ge\pi$. We need to find the region within $\text{A}>0$, $\text{B}>0$, $\text{A}+\text{B}<\pi$ where this holds.

We can now normalize the vaiables. Normalize the angles by dividing by $\pi$. Let $\displaystyle\text{a}=\frac{\displaystyle\text{A}}{\displaystyle\pi}$, $\displaystyle\text{b}=\frac{\displaystyle\text{B}}{\pi}$, and $\displaystyle\text{c}=\frac{\displaystyle\text{C}}{\displaystyle\pi}$. Since $\text{A}+\text{B}+\text{C}=\pi$, we have: $\text{a}+\text{b}+\text{c}=1$ and $\text{c}=1-\text{a}-\text{b}$, with: $\text{a}>0$, $\text{b}>0$, and $\text{c}>0\space\Longrightarrow\space1-\text{a}-\text{b}>0\space\Longrightarrow\space\text{a}+\text{b}<1$. The region $\text{a}>0$, $\text{b}>0$, and $\text{a}+\text{b}<1$ is a right triangle in the $\text{a}$-$\text{b}$ plane with vertices $(0,0)$, $(1,0)$, and $(0,1)$, and area:

$$\text{Area}=\frac{1}{2}\cdot1\cdot1=\frac{1}{2}\tag2$$

Substitute into the inequality: $\left(\pi\text{a}\right)x+\left(\pi\text{b}\right)y\ge\pi\left(1-\text{a}-\text{b}\right)$. Divide through by $\pi$: $\text{a}x+\text{b}y\ge1-\text{a}-\text{b}$, $\text{a}x+\text{b}y+\text{a}+\text{b}\ge1$, and $\text{a}\left(x+1\right)+\text{b}\left(y+1\right)\ge1$. The probability becomes the area where $\text{a}\left(x+1\right)+\text{b}\left(y+1\right)\ge1$ within this triangle, divided by $\displaystyle\frac{\displaystyle1}{\displaystyle2}$.

We can now compute the probability. Consider the line $\text{a}\left(x+1\right)+\text{b}\left(y+1\right)=1$. Intersection with $\text{a}=0$: $0\cdot\left(x+1\right)+\text{b}\left(y+1\right)=1\space\Longrightarrow\space\displaystyle\text{b}=\frac{\displaystyle1}{\displaystyle y+1}$. Since $y>0$, $y+1>1$, so $0<\frac{\displaystyle1}{\displaystyle y+1}<1$, and $\text{a}+\text{b}<1$ holds since $\text{b}<1$. Point $\left(0,\frac{\displaystyle1}{\displaystyle y+1}\right)$. Intersection with $\text{b}=0$: $\text{a}\left(x+1\right)=1\space\Longrightarrow\space\text{a}=\frac{\displaystyle1}{\displaystyle x+1}$. Since $x>0$, $0<\frac{\displaystyle1}{\displaystyle x+1}<1$. Point $\left(\frac{\displaystyle1}{\displaystyle x+1},0\right)$. The line connects $\left(0,\frac{\displaystyle1}{\displaystyle y+1}\right)$ and $\left(\frac{\displaystyle1}{\displaystyle x+1},0\right)$, both within the triangle’s boundaries. Test $(0,0)$: $0\cdot\left(x+1\right)+0\cdot\left(y+1\right)=0<1$, so $\text{a}\left(x+1\right)+\text{b}\left(y+1\right)<1$ inclues $(0,0)$, and $\ge1$ is the region “above” the line toward $\text{a}+\text{b}=1$. This is the triangle with vertices $(0,0)$, $\left(0,\frac{\displaystyle1}{\displaystyle y+1}\right)$ and $\left(\frac{\displaystyle1}{\displaystyle x+1},0\right)$: base along $\text{a}$-axis: $\frac{\displaystyle1}{\displaystyle x+1}$, height along $\text{b}$-axis: $\frac{\displaystyle1}{\displaystyle y+1}$. So the area gives:

$$\text{Area}=\frac{1}{2}\cdot\frac{\displaystyle1}{\displaystyle x+1}\cdot\frac{\displaystyle1}{\displaystyle y+1}\tag3$$

So the area where $\text{a}\left(x+1\right)+\text{b}\left(y+1\right)\ge1$, gives:

$$\frac{1}{2}-\frac{1}{2}\cdot\frac{\displaystyle1}{\displaystyle x+1}\cdot\frac{\displaystyle1}{\displaystyle y+1}=\frac{1}{2}\left(1-\frac{1}{\left(x+1\right)\left(y+1\right)}\right)\tag4$$

So, the probability is:

$$\mathbb{P}=\frac{\text{area where inequality holds}}{\text{total area}}=\frac{\displaystyle\frac{\displaystyle1}{\displaystyle2}\left(1-\frac{\displaystyle1}{\displaystyle\left(x+1\right)\left(y+1\right)}\right)}{\displaystyle\frac{\displaystyle1}{\displaystyle2}}=1-\frac{1}{\left(x+1\right)\left(y+1\right)}\tag5$$

Answer 2

Let $x = \frac{A}{\pi}$, $y = \frac{B}{\pi}$, and $z = \frac{C}{\pi}$ (not to be confused with your use of $x$ and $y$). Then (x, y, z) ~ Dirichlet(1, 1, 1). An equivalent geometric formulation is that $(x, y, z)$ is uniformly distributed on the triangle formed by the points $(1, 0, 0)$, $(0, 1, 0)$, and $(0, 0, 1)$, which lie on the plane $z = 1 - x - y$.

The three edges of the triangle can be parametrized as $(t, 1-t, 0)$, $(t, 0, 1-t)$, and $(0, t, 1-t)$, where $0 \le t \le 1$.

Let $\alpha$ and $\beta$ be arbitrary positive real numbers. Then the condition $A\alpha + B\beta > C$ is equivalent to $z < \alpha x + \beta y$.

The intersection of $z = 1 - x - y$ and $z = \alpha x + \beta y$ is $y = \frac{1 - (\alpha + 1) x}{\beta + 1}$. This plane intersects the edges of the triangle at the points $\left(\frac{1}{\alpha + 1}, 0, \frac{\alpha}{\alpha + 1}\right)$ and $\left(0, \frac{1}{\beta + 1}, \frac{\beta}{\beta + 1} \right)$.

So, the region where $z$ is greater than $\alpha x + \beta y$ is the triangle with vertices at those two points and $(0, 0, 1)$. The side lengths of the triangle are $\frac{\sqrt{2(\alpha^2 + \beta^2 + \alpha + \beta - \alpha\beta + 1)}}{(\alpha + 1)(\beta + 1)}$, $\frac{\sqrt{2}}{\alpha + 1}$, and $\frac{\sqrt{2}}{\beta + 1}$. The semiperimeter is $\frac{\alpha + \beta + 2 + \sqrt{\alpha^2 + \beta^2 + \alpha + \beta - \alpha\beta + 1}}{\sqrt{2}(\alpha + 1)(\beta + 1)}$. By Heron's formula (and a lot of algebra), the area of the triangle is $\frac{\sqrt{3}}{2(\alpha + 1)(\beta + 1)}$.

But the support triangle as a whole is an equilateral triangle with a side length of $\sqrt{2}$, which has an area of $\frac{\sqrt{3}}{2}$.

So $P(z > \alpha x + \beta y) = \frac{\frac{\sqrt{3}}{2(\alpha + 1)(\beta + 1)}}{\frac{\sqrt{3}}{2}} = \frac{1}{(\alpha + 1)(\beta + 1)}$.

So $P(A\alpha + B\beta > C) = P(z < \alpha x + \beta y) = 1 - \frac{1}{(\alpha + 1)(\beta + 1)}$.

Or, replacing $\alpha$ and $\beta$ with $x$ and $y$,

$$\boxed{P(Ax + By > C) = 1 - \frac{1}{(x + 1)(y + 1)}}$$

Q.E.D.